For a function $g$, we are given that $g(-6)=-8$ and $g'(-6)=3$. What's the equation of the tangent line to the graph of $g$ at $x=-6$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y+6=-8(x-3)$ (Choice B) B $y-3=-8(x+6)$ (Choice C) C $y+6=3(x+8)$ (Choice D) D $y+8=3(x+6)$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $g'(-6)$ gives the slope of the tangent line to the graph of $g$ where $x=-6$. We are given that $g'(-6)=3$, so the slope of the tangent line is $3$. Furthermore, we are given that $g(-6)=-8$, which means the point of intersection of the tangent line and the graph is $(-6,-8)$. To summarize, the tangent line has a slope of $3$ and it passes through the point $(-6,-8)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-(-8)&=3(x-(-6)) \\\\ y+8&=3(x+6) \end{aligned}$ The equation is $y+8=3(x+6)$.